What is the derivative of $y=x^lnx$ ?

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Answer 1 · 2015-12-16T23:35:30

$(2x^(lnx)lnx)/x$

Take natural log of both sides

$lny=lnx^(lnx)$

Rewrite write hand side using properties of logarithms

$lny=lnx(lnx)$

$lny=(lnx)^2$

Differentiate both sides with respect to $x$

$1/y(dy)/dx=2lnx(1/x)$

$1/y(dy)/dx=(2lnx)/x$

Multiply both sides by $y$

$(dy)/dx=y((2lnx)/x)$

Note that $y=x^(lnx)$ so we write

$(dy)/dx=x^(lnx)((2lnx)/x)$

$(dy)/dx=(2x^(lnx)lnx)/x$

What is the derivative of y=x^lnxy=x^lnx?